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Let $U=\{1,2,3,4,5,6,7,8,9\}, A=\{1,2,3,4\}$, $B=\{2,4,6,8\}$ and $C=\{3,4,5,6\}$. Find,
(i) $A^{\prime}$,
(ii) $B^{\prime}$
(iii) $(A \cup C)^{\prime}$
(iv) $(A \cup B)^{\prime}(\mathrm{v})\left(A^{\prime}\right)^{\prime}(\mathrm{vi})(B-C)^{\prime}$
(i) $A^{\prime}$,
(ii) $B^{\prime}$
(iii) $(A \cup C)^{\prime}$
(iv) $(A \cup B)^{\prime}(\mathrm{v})\left(A^{\prime}\right)^{\prime}(\mathrm{vi})(B-C)^{\prime}$
Solution:
2983 Upvotes
Verified Answer
Here, $U=\{1,2,3,4,5,6,7,8,9\}$
$A=\{1,2,3,4\}$
$B=\{2,4,6,8\}$
and $C=\{3,4,5,6\}$
(i) $A^{\prime}=U-A=\{5,6,7,8,9\}$
(ii) $B^{\prime}=U-B=\{1,3,5,7,9\}$
(iii) $\mathrm{A} \cup \mathrm{C}=\{1,2,3,4\} \cup\{3,4,5,6\}$
$=\{1,2,3,4,5,6\}$
Hence $(A \cup C)^{\prime}=\mathrm{U}-(A \cup C)=\{7,8,9\}$
(iv) $A \cup B=\{1,2,3,4\} \cup\{2,4,6,8\}$
$=\{1,2,3,4,6,8\}$
Hence $(A \cup B)^{\prime}=U-(A \cup B)=\{5,7,9\}$
(v) $A^{\prime}=U-A=\{5,6,7,8,9\}$
Hence $\left(A^{\prime}\right)^{\prime}=U-A^{\prime}=\{1,2,3,4\}$
(vi) $B-C=\{2,4,6,8\}-\{3,4,5,6\}=\{2,8\}$
Hence $(B-C)^{\prime}=U-(B-C)=\{1,3,4,5,6,7,9\}$
$A=\{1,2,3,4\}$
$B=\{2,4,6,8\}$
and $C=\{3,4,5,6\}$
(i) $A^{\prime}=U-A=\{5,6,7,8,9\}$
(ii) $B^{\prime}=U-B=\{1,3,5,7,9\}$
(iii) $\mathrm{A} \cup \mathrm{C}=\{1,2,3,4\} \cup\{3,4,5,6\}$
$=\{1,2,3,4,5,6\}$
Hence $(A \cup C)^{\prime}=\mathrm{U}-(A \cup C)=\{7,8,9\}$
(iv) $A \cup B=\{1,2,3,4\} \cup\{2,4,6,8\}$
$=\{1,2,3,4,6,8\}$
Hence $(A \cup B)^{\prime}=U-(A \cup B)=\{5,7,9\}$
(v) $A^{\prime}=U-A=\{5,6,7,8,9\}$
Hence $\left(A^{\prime}\right)^{\prime}=U-A^{\prime}=\{1,2,3,4\}$
(vi) $B-C=\{2,4,6,8\}-\{3,4,5,6\}=\{2,8\}$
Hence $(B-C)^{\prime}=U-(B-C)=\{1,3,4,5,6,7,9\}$
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